A process that drills holes in crankshafts that are subsequently ground has been

Question

A process that drills holes in crankshafts that are subsequently ground has been found to be in good statistical control, with the following statis- tics used to estimate the mean and the variance of the hole diameter: x 15.050 millimeters, F-0.110 millimeter. It is also known that 95% of the hole diameters lie within the range 14.945 to 15.155 millimeters. Find the size of the samples employed to produce the ranges R used to estimate the standard deviation of the process. Assume that the hole diameters are normally distributed.

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then,

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be found using Excel Function……………………..(3)

Property of symmetry: P(Z > t) = P(Z < - t)………………………………………….(4)

In the context of Control Charts, µ and are respectively replaced by their sample estimates, Xdouble bar and Rbar/d2 ………………………………….…………………………(5)

where d2 is a constant available in standard Control Charts Constants Table, whose value depends on the sample size, n.

Now, to work out the solution,

Let X = hole diameter. We are given that X ~ N(µ, 2), where vide (5) above µ = 15.050 (given), = 0.110/d2 [given]……………………………………………………………..(6)

Also given is that 95% of the holes have diameter between 14.945 and 15.155, which when translated into probability language would mean:

P(14.945 < X < 15.155) = 0.95

=> vide (2) and (6 )above,

P[{(14.945 – 15.050)/( 0.110/d2)} < Z < {(14.945 – 15.155)/( 0.110/d2)}] = 0.95

=> P[{- (0.105)/( 0.110/d2)} < Z < {(0.105)/( 0.110/d2)}] = 0.95

=> {(0.105)/( 0.110/d2)} = 1.96 [because of symmetry property of Normal Distribution – vide (4) above and from Excel Function on Standard Normal Distribution P(|Z| < 1.96) = 0.95]

=> d2 = 2.052

From Control Charts Constants Table, we note that d2 = 2.059 for n = 4.

Thus the sample size employed is 4. ANSWER

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